A 35.0 g ball at the end of a string is swung in a vertical circle with a radius of 35.0 cm. The rotational velocity is 212.0 cm/s.
(a)Find the tension (in N) in the string at the top of the circle.
(b)Find the tension (in N) in the string at the bottom of the circle.
After a mishap, an 82.8 kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here.
(angle for T1 = 15)
(angle for T2 = 10)
Calculate the tension (in N) in the first rope,T1, if the person is momentarily motionless. (Enter the magnitude.)
Calculate the tension (in N) in the second rope, T2, if the person is momentarily motionless. (Enter the magnitude.)
A 35.0 g ball at the end of a string is swung in a vertical circle with a radius of 35.0 cm. The rotational velocity is 212.0 cm/s.
(a)Find the tension (in N) in the string at the top of the circle.
Tension at top: mv^2/r-mg
Tension at bottom: mv^2/r+mg
I can't visualize the second problem without the figure.
To find the tension in the string at the top and bottom of the circle, we can make use of the centripetal force equation.
(a) Tension at the top of the circle:
At the top of the circle, the tension in the string provides the centripetal force required to keep the ball moving in a circular path. The tension force acts downward, opposing the force of gravity.
To find the value of the tension at the top, we need to equate the centripetal force to the sum of the gravitational force and the tension force:
T + mg = (mv^2) / r
Where:
T = Tension in the string
m = Mass of the ball (35.0 g or 0.035 kg)
g = Acceleration due to gravity (9.8 m/s^2)
v = Rotational velocity (212.0 cm/s or 2.12 m/s)
r = Radius of the circle (35.0 cm or 0.35 m)
Let's solve this equation to find the tension at the top:
T + (0.035 kg)(9.8 m/s^2) = (0.035 kg)(2.12 m/s)^2 / 0.35 m
T + 0.343 N = 0.432 N
Subtracting 0.343 N from both sides:
T = 0.432 N - 0.343 N
T = 0.089 N
Therefore, the tension in the string at the top of the circle is 0.089 N.
(b) Tension at the bottom of the circle:
At the bottom of the circle, the tension in the string still provides the centripetal force required to keep the ball moving in a circular path. However, at this point, the tension force acts upward, aiding the force of gravity.
Using the same equation as before, we can find the tension at the bottom:
T - mg = (mv^2) / r
T - (0.035 kg)(9.8 m/s^2) = (0.035 kg)(2.12 m/s)^2 / 0.35 m
T - 0.343 N = 0.432 N
Adding 0.343 N to both sides:
T = 0.432 N + 0.343 N
T = 0.775 N
Therefore, the tension in the string at the bottom of the circle is 0.775 N.